Solve 3^x-2*3^x*3^-2=7

$${ 3 }^{ x } -2 \times { 3 }^{ x } \times { 3 }^{ -2 } =7$$

Solve for x

$$x=2$$

Steps Using Definition of Logarithm

Calculate 3 to the power of -2 and get \frac{1}{9}.

$$3^{x}-2\times 3^{x}\times \frac{1}{9}=7$$

Multiply 2 and \frac{1}{9} to get \frac{2}{9}.

$$3^{x}-\frac{2}{9}\times 3^{x}=7$$

Combine 3^{x} and -\frac{2}{9}\times 3^{x} to get \frac{7}{9}\times 3^{x}.

$$\frac{7}{9}\times 3^{x}=7$$

Multiply both sides by \frac{9}{7}, the reciprocal of \frac{7}{9}.

$$3^{x}=7\times \frac{9}{7}$$

Multiply 7 and \frac{9}{7} to get 9.

$$3^{x}=9$$

Take the logarithm of both sides of the equation.

$$\log(3^{x})=\log(9)$$

The logarithm of a number raised to a power is the power times the logarithm of the number.

$$x\log(3)=\log(9)$$

Divide both sides by \log(3).

$$x=\frac{\log(9)}{\log(3)}$$

By the change-of-base formula \frac{\log(a)}{\log(b)}=\log_{b}\left(a\right).

$$x=\log_{3}\left(9\right)$$

Solve for x (complex solution)

$$x=\frac{2\pi n_{1}i}{\ln(3)}+2$$

$$n_{1}\in \mathrm{Z}$$